will it take off?

[Paddy]

Patrick G,

All problems do have assuptions as you and I know from our sience/engineering back grounds. The entire difference between the fly and no-fly camp falls on the context of the convayer matching the speed of the plane. You summerize this in your statement;

""Now, with these assumption made, the only way for our plane to take off and remain flying is for it to travel, or move, from the start of the MCB toward the opposite end of the MCB, all the while gaining speed until it reaches 200 mph with relation to the ground. If it remains in one place on the MCB, I think we can assume it will not take off and continue to fly."

This where I think your assumptions begin to fail. Your statement;

"I am not reproducing the rest of your argument because it does not matter since it assumes the plane can move forward and it can't."

The problem clearly states, "as the plane moves''

Your next statement really starts to deviate from the problem;

"I will try to explain how the MCB can counter the thrust of those powerful Binford Belchfire 5000 Tim The Tool Man Taylor signature model engines!

I ask a simple question, where in the problem does it state or even imply the convayer can counter a plane's thrust? That is flat out re-wording the problem. As the plane moves..the conveyor has a system that tracks the speed of the plane and matches it exactly"" If you re-word the problem and make a new problem, sure it would take magic for this convayer but it could counter the small amount of friction. But than wasn't the in the problem.

You continue with this re-wrighting the problem;

"In your response you said something aboiut the friction not being too great. Well that depends on the speed of the MCB doesn't it? The faster you run the wheels the more force is transfered due to friction. Since there is no upper limit to the speed of the MCB, it runs as fast as is required to concel the thrust of the engines."

What do you mean there is no upper limit of the convayer?? The problem clearly states it is matched but in oposite direction. And again you move back to thrust but this time use the word concealed. Where do you get all these additional terms?

You state;
How did the plane suddenly get to be going 10MPH???
Wouldn't it have to go 9MPH first and 8MPH before that and 7,6,5,4,3,2,1,,,0.0000001MPH before that? The MCB is a very good device and doesn't wait for the A/C to get to 10MPH before it starts to work. The MCB starts countering the planes attempt to move

Of cource the plane doesn't instantly start at 10 mi/hr, it would begin gradually. So what the point? But you do it again in your statement the 'MCB starts countering the planes attemp to move' Theres two faults here 1) there's no attempt, problem states "as the plane moves", 2) the MCB doesn't counter the planes attempt to move it MATCHes the planes speed.


What tops it off is your statement implies the "Will fly" camp needs to use assumptions tt make the point.

Let me say it again. THE PROBLEM WAS WORDED TOO LOOSELY and it requires that unstated assumptions be made to support either camp.

What assumption do you think the Will-Fly camp is using that you think is out?
I think I know, we think the convayer matches the planes speed and the No-fly think it matches the wheel speed. In your camp it all comes down to convayer moving at high rates of speed and plane sitting still a top of spinning wheels. All the while using terms and arguments like matching thrust
or countering friction.

 

[jk96]

Paddy,

Thought this was kind of funny. Someone finally got tired of all the no flyers and videotaped a very simple experiment

YouTube - Airplane Conveyer Belt Experiment

jk

 

[RayH]

Lots of talk and long posts and I still think its a very simple problem.
The conveyor will only go as fast as the plane is capable of going (It matches the planes speed). A plane is limited by its thrust and drag. for instance, a Cessna 172 can only go about 200, a 727 can only go about 600, an F15 can only go about 1500. The conveyors speed can not go faster backwards than (insert plane type) can go forward.
This is so simple for me and I dont understand how it can be disputed.

The only arguement against this is that the question meant to state that the conveyor matches the wheels speed. If thats the case, guess what, it will always match wheel speed, it doesnt matter if the plane is moving or not. The way that I know the conveyor and wheels are ALWAYS matching each other is because the wheels are not skidding on the conveyor. If the wheels are not skidding, the conveyor and wheels are offsetting each other at the same speed. This also seems so simple to me I cant see how it can be disputed.

 

[patrick_g]

OK, OK, IT seems someone realy wants me to comment. OK, here is a comment.

If you go with the versioin where the speed of the conveyor matches the speed of the airplane then the plane will most assuredly take off. Lets say the plane needs 100KTS IAS and there is no wind except as caused by drag on the survface of the MCB and the propwash.

A speedometer attached to the landing gear will register 200KTS the IAS will be 100 and the A/C will rotate and climb out mormally if the tires don't explode from running 200 KTS. This is the obvious conclusion from the afore mentioned interpretatioin. It is quite mundane and pedestrian, not much fun and sure doesn't require a rocket scientist to understand.

The "other version" was just a heck of a lot more fun and would have been more fun yet if more folks could have followed the argument instead of insisting on all sorts of ridiculous assumptions fraught with inapplicability.

Pat (an equal opportunity debater)

 

[jk96]

Pat,

Welcome to the "fly side" - lol

I retract my earlier statement. You do in fact "get it". Obviously much better than I do. Your ability to prove your theory on why it would not move was in fact very entertaining, and all admit, well above my knowledge base. Now I'm going to go watch some ice melt as well. Or at least have some fun trying to bust it up with my tractor - I almost forgot, tractors were the main topic on this website weren't they?

jk

 

[Egon]

JK; Pat has been waffling. He's also having a lot of fun!

Next week he has to clean out his fish ponds! and get ready for next years stock.

 

[patrick_g]

Turnkey, I'm curious, how is it that you declare the friction of the wheels can't equal the thrust? The force produced by the friction is propoertional to speed. With enough speed you can generate any finite force needed. Of course in the real world the bearings will melt and the wheels will fly apart but this isn't the real world.

Please note that all the force required to equal the thrust of the plane's engine is NOT produced by wheel friction. The rotating assembly also has inertia and stores energy when spun up. in order to get that energy into the rotating mass yoiu have to apply a force. That force is supplied by the friction between the tire and the MCB. Now if you claim the coefficient of friction is too small to transfer that force I have to agree that in the real world that is a consideration but in this world of MAGIC conveyor belts it should not be a problem (this particular magic conveyor belt has a very high traction coefficient.)

Now for JK and Tom K (Thanks again Tom for vote of confidence) Now as regards the relativistic issue. The speed of light is not approached in a linear fashion. The energy required to accelerate a mass (even something as light as a single electron) to the speed of light is infinite. The mass of the accelerated object increases such that it would be infinite at the speed of light. (See Einstein, A.)

This supports the contention that the MCB could accelerate as required to hold the plane back for a very long time even several refuelings (forever theoretically.) Of course the MCB will be running way faster than the motion of the planes inertial navigator is indicating. Requires the "OTHER" assumption.

There are two basic solutions to the problem:

1. It takes off with the wheels spining twice as fast as the plane is flying because the MCB was going the same speed as the plane.

2. It never takes off because the MCB can continue to accelerate at the rate required to generate through frictioin and inertia of the wheels a force equal to the plane's thrust.

2.1 Extra frills are things like the air pumped by contact with the whirling dervish of a MCB will allow the plane to generate lift and lift off. It would then accelerate and may or may not stall (we don't have enough information) when it got to the end of the MCB.)

2 and 2.1 are very much more interesting cases than 1 but both 1 and 2 are valid dep[ending on interpretation of the problem statement.

I'm not sure but I think I was insulted by some miscreant who may have compared me to Slick Willy. Just because very precise but small differences are teased out of a statement does not impart the sleaze factor of Slick Willy to the teaser. If guilt by associaton that tenuous is allowed then we are all breathing the same air as ****** and Sadam so where does that leave us?????

Tom K. Please consider any physics stuff I injected as a tax rebate. I was a physics major and math minor under the Viet Nam GI Bill. Thanks to all who paid federal taxes in the late 60's and early 70's. I later found true religion (COMPUTERS) and got a BS in Comp Sci and a MS in Software Engineering before getting into the engineering of training solutions (MS in Instructional Technology) Explains my professorial approach to problems like this. It does not explain my perverse sense of humor that frequently goes not only unappreciated but unnoticed.

Pat

 

[turnkey4099]

Turnkey, I'm curious, how is it that you declare the friction of the wheels can't equal the thrust? The force produced by the friction is propoertional to speed. With enough speed you can generate any finite force needed. Of course in the real world the bearings will melt and the wheels will fly apart but this isn't the real world.

Please note that all the force required to equal the thrust of the plane's engine is NOT produced by wheel friction. The rotating assembly also has inertia and stores energy when spun up. in order to get that energy into the rotating mass yoiu have to apply a force. That force is supplied by the friction between the tire and the MCB. Now if you claim the coefficient of friction is too small to transfer that force I have to agree that in the real world that is a consideration but in this world of MAGIC conveyor belts it should not be a problem (this particular magic conveyor belt has a very high traction coefficient.)

<snip>

Pat

I base the lack of enough friction on real world conditions. As the problem is stated, assuming a 100 kts TO speed, the weels would be going 200 kts. Not even close to developing anough friction, rolling resisstance or rotational intertia to soad up even a small engines thrust.

So far no-one has come up with a reason that the plane would not begin to move when the engine wound up. YOu might argue the 'magic' thing after it gets going (unusuccessfully IMO) but how do you get enough friction to keep it from going from say 0mph to 1 mph? The problem clearly states that the plane 'moves'.

Harry K

 

[Tom_Veatch]

...
So far no-one has come up with a reason that the plane would not begin to move when the engine wound up. YOu might argue the 'magic' thing after it gets going (unusuccessfully IMO) but how do you get enough friction to keep it from going from say 0mph to 1 mph? The problem clearly states that the plane 'moves'.

Harry K

But, it doesn't say what it moves in relation to. A person sitting under a tree beside the conveyor sees the plane move at a speed relative to the ground. Someone sitting on the conveyor sees the plane move at a speed relative to the conveyor. The problem doesn't explicitly define which of those relative speeds is matched by the conveyor.

In fact, the answer to the problem is entirely contained in that definition of speed. If the tree sitter sees movement, the conveyor is matching the plane's speed relative to the ground and the plane flies. If only the conveyor sitter sees movement, the conveyor is running at a speed relative to the ground that's the same as the plane's speed relative to the conveyor.

Since the problem statement doesn't define which aircraft speed is matched, either one can be assumed. And, depending on which assumption is made, either answer can be theoretically justified using the theories of Newtonian physics. To apply those theories to physical objects, you have to move from the scientific/theoretical realm into the engineering realm. Now, you have a whole mess of new assumptions that have to be made (and stated).

I believe a successful defense of the "no-fly" case ultimately depends on the nature of time. Is time truly a continuous function or does it have a quantum nature. We know Newtonian physics is a special case only valid at speeds that are slow relative to the speed of light. Is it also a special case which is invalid over extremely small units (quanta) of time?

...
The force produced by the friction is propoertional to speed.
...

Pat, I believe you are thinking of viscous forces which are proportional to speed. Friction force is proportional to the normal or perpendicular force between surfaces and, other than different values for static and dynamic coefficients of friction, are not generally dependent on the relative velocities of the surfaces. Of course, you may be assuming a fluid layer of lubrication, e.g. grease, oil, molten metal, with no solid surface contact between the bearing races. In that case you are correct. The bearing force will be viscous in nature. But, you also need to account for viscosity changes in the lubrication as the bearing assembly gets hotter.

In any case, I believe the bearing force can be neglected as small compared to the force necessary to accelerate the tire/wheel rotation. I've been tempted to calculate the conveyor acceleration necessary to hold an airplane static relative to the tree sitter. But, so far, have successfully resisted that temptation.

...
I still want to know if you have a bird in a box and know the weight of the bird and box separately whether or not you can tell with a pair of scales whether the bird is perching or flying.

At equilibrium in each state, the box containing the flying bird and the box containing the perched bird will weigh the same on a set of scales external to the box. The Laws of Thermodynamics admit no other solution. A more interesting question is what happens to the scales at the instant the bird lands or takes off, and a comparison of those two events. I'm having a hard time not seeing a violation of the 2nd Law of Thermodynamics in those two events.

 

[HTWT]

I want to know how to find the three points of intersection of the 2 equations

1. Y= 2^x
2. Y= X^2

Two solutions are easy by inspection but what about the negative value for X.

 

[patrick_g]

Tom, My A/C all use lubricants in the wheel bearings. Besides friction is a minor contributor compared to the rotating mass energy storage that alows the MCB to accelerate at a constant rate to produce the force equivalent to the propulsion systems thrust. Folks just kept on focusing on friction, due to familiarity I suppose.

If you are in for a penny you are in for a pound. If the bird in flight and at rest show the same weight on the scales...

Lets make it a very tall hour glass with very course sand. Will a sensitive scale show the big grains hitting the bottom? Either group the must fly crowd or the heredical no fly bunch would realize that the hour glass would weigh the same after the sands of time had shifted all to the bottom no matter if the red end or the black one was on bottom when the sand stopped. The question is does the weight change while the grains are trickling down? At the start with a freshly inverted hour glass does it weigh more, less, or the same. What becomes of the potential energy which the elevated sand had. Does the sensitive scale show any variatioin as grains free fall and then come to rest?

Can a man in an enclosed elevator car tell if the elevator is accelerating upward as opposed to a large mass appearing under the car?

Can there be a doubter doubting that there could be a doubter doubting?

Can two mediteranean larks working in tandom carry a cocoanut? (appologies to Monty Python)

Time and about everything else looks smoothly continuous when viewed from the right distance. Conversely, just about everything including time starts looking as granular as a low res computer graphic on sufficiently close inspection. "Great fleas have little fleas upon their backs to bite 'em,: And little fleas have lesser fleas, and so ad infinitum.

Pat

 

[MossRoad]

So the plane powers up and moves forward. The MCB powers up and moves backwards. At this point the plane isn't standing still, is it? ...

Yes.

At the very moment the plane tries to move forward .00000001mm, the conveyor moves backwards .00000001mm, right? Now instead of .00000001mm run this decimal place out to a bazillion places to the right. Every atom width movement forward would be countered by an atom width movement backwards. If the conveyor could truly match the forward speed of the plane in the opposite direction, you would hear the engine run up but would not see the plane or the conveyor move, just as if the brakes were on.

 

[Tom_Veatch]

I want to know how to find the three points of intersection of the 2 equations

1. Y= 2^x
2. Y= X^2

Two solutions are easy by inspection but what about the negative value for X.

Since you are talking about points of intersection, I assume we are working in 2-space with the 3rd coordinate held to a constant value. IOW, an X-Y plane. I further assume we are working in the real instead of the complex number system.

That being the case, I only see one obvious solution, the point (2,4).

I question there being three points of intersection under the assumption stated above.

Eq. 1 is an exponential curve asymptotic to the lines X= +inf. and Y= 0. Equation 2 is a parabola, concave upward with a vertex at (0,0), asymptotic to the lines X = +inf and X= -inf.

There are only two points short of the asymptotes at which these curves can intersect, the obvious point (2,4) in the 1st quadrant and, in the 4th quadrant:

X = (2/ln(2)) * ln(abs(X)) where X < 0

This expression for the X-value of an intersection point can be obtained by simple algebraic manipulation but is a transcendental equation which cannot be solved by simple algebraic methods. An approximate numerical value can be obtained by straightforward numeric computational techniques but I suspect that an exact solution can only be expressed as a slowly converging infinite series similar to the case for other transcendental values and functions.

I suppose, since both curves are asymptotic in the 1st quadrant to the line X= +inf, one could claim (+inf, +inf) as a third intersection point. But that claim would be made in the same sense that parallel lines are claimed to intersect at infinity. This implies that "infinity" is an actual rather than a conceptual value, so I argue that there are only two actual intersection points on the real plane.

If you admit complex solutions, X = a + bj where j = sqrt(-1), then perhaps one, or even many, complex roots exist, but the half-life of my expertise with complex numbers has expired many times over so I'll pass on that one.

Now, to get on topic for this forum. I have a 22HP CUT. What size moldboard plow would be appropriate for that size tractor and how long do you think it might take to plow the 1st quadrant of the X-Y plane?

 

[Tom_Veatch]

Tom, My A/C all use lubricants in the wheel bearings. Besides friction is a minor contributor compared to the rotating mass energy storage that alows the MCB to accelerate at a constant rate to produce the force equivalent to the propulsion systems thrust. Folks just kept on focusing on friction, due to familiarity I suppose.

Yeah, I was just being a little too nit-picking. Besides, if there were no lubricants in the bearing, they would soon be lubricated by molten metal anyway.

Lets make it a very tall hour glass with very course sand. Will a sensitive scale show the big grains hitting the bottom? Either group the must fly crowd or the heredical no fly bunch would realize that the hour glass would weigh the same after the sands of time had shifted all to the bottom no matter if the red end or the black one was on bottom when the sand stopped. The question is does the weight change while the grains are trickling down? At the start with a freshly inverted hour glass does it weigh more, less, or the same. What becomes of the potential energy which the elevated sand had. Does the sensitive scale show any variatioin as grains free fall and then come to rest?

Let's simplify the configuration a little. Assume there is only one grain of sand in the hourglass. If we can conclude something about that configuration, the conclusion extends to the case of many grains. Further assume that a perfect vacuum exists within the hourglass so we can neglect any aerodynamic influences inside the hourglass. Agreed?

At the start of the experiment, the grain, which weighs G pounds, is somehow restrained in the upper chamber. Perhaps the opening is slightly too small to allow passage. The hourglass, containing the grain, is resting on a sensitive, instantaneously registering scale. The hourglass alone, without the grain, has been previously determined to weigh H pounds. Thus the total system, hourglass plus restrained grain weighs H + G pounds.

The experiment begins by, say heating the hourglass so that the opening expands enough to allow the grain to pass through and fall to the bottom of the lower chamber. As soon as the grain begins to fall, the weight of the grain is no longer being supported by the hourglass and measured on the scale. Therefore, during the period of free fall the scale will register only the weight of the hourglass, H pounds.

The potential energy, PE = Gh, of the grain in the upper chamber is converted to kinetic energy, KE = 0.5 (G/g) * V^2. The PE decreases as the height (h) of the grain above the bottom of the hourglass decreases. The KE increases as the square of the speed of the grain's fall increases. At any point during the fall, the sum, PE + KE, will be constant and equal to the initial potential energy.

When the grain impacts the bottom of the hourglass, the weight of the grain will again be supported by the hourglass and the scale will again register H + G. So the scale will register H+G before the fall, H during the fall, and H + G once the grain comes to rest at the bottom of the fall.

So what happened to the kinetic energy of the grain that it had accumulated at the instant prior to impact. Any number of things. But if we assume neither the hourglass or the grain was physically deformed that energy will heat both grain and hourglass so that both are very slightly warmer than they were at the instant before the fall. If we allow deformation of either grain or hourglass bottom, some of the energy will be used by the work to accomplish that deformation and the amount of heating will be slightly less.

I believe that touches all the points in your question except to say that what happens to one grain happens to all grains in the falling stream. So, once the stream of sand starts falling, the scale will detect only the weight of the hourglass plus only that sand that is not falling.

Can a man in an enclosed elevator car tell if the elevator is accelerating upward as opposed to a large mass appearing under the car?

According to General Relativity an observer cannot distinguish between the two conditions in an inertial reference frame. However, the surface of the earth is a rotating sphere and is not a true inertial reference system. Close enough to be considered one in cases of airplanes flying off conveyor belts, but not in the strictest sense. That fact has to be taken into account for every launch from Vandenburg or Kennedy, or for any launch from the Earth's surface except an exactly vertical launch from either of the Earth's poles.

Therefore, the elevator moving upward perpendicular to the Earth's surface experiences a very slight Coriolis acceleration in a Westerly direction. That acceleration would be absent if the increase in apparent weight were due to the sudden appearance of a large concentrated mass below the elevator. Therefore, with sufficiently sensitive instrumentation, the passenger could determine the difference between the two cases.

Can there be a doubter doubting that there could be a doubter doubting?

Can two mediteranean larks working in tandom carry a cocoanut? (appologies to Monty Python)

Wow! This is beginning to remind me of the my Master's Degree Orals. I'm going to pass on those two.

Time and about everything else looks smoothly continuous when viewed from the right distance. Conversely, just about everything including time starts looking as granular as a low res computer graphic on sufficiently close inspection. "Great fleas have little fleas upon their backs to bite 'em,: And little fleas have lesser fleas, and so ad infinitum.

Pat

From my readings several years ago, I knew there were some theoretical physicists working on one or more of the GUTs who were hypothesizing a granularity or quantum nature of time. I had not heard or read that those hypotheses were in general acceptance or had reached the status of theory. I wonder how it could be verified experimentally.

Well, to get back on topic, will allowing my brush-hog to free fall from a full elevation on my 3pt hitch relieve some of the load on my rear tires? Or will the Coriolis forces during the fall bend the lift arms if the tractor happens to be facing North or South when I do it?

 

[Tom_Veatch]

Yes.

At the very moment the plane tries to move forward .00000001mm, the conveyor moves backwards .00000001mm, right? Now instead of .00000001mm run this decimal place out to a bazillion places to the right. Every atom width movement forward would be countered by an atom width movement backwards.
...

What is your explanation for the method by which the movement of the conveyor is transmitted to the airframe?

 

[MossRoad]

What is your explanation for the method by which the movement of the conveyor is transmitted to the airframe?

The entire weight of the aiplane is on the little patch of rubber tires that is contacting the conveyor. None of it will ever get transferred to the wings because it will never move forward one spec.

Put a brick on a treadmill and slowly turn on the treadmill. The brick moves backwards. Now put a roller skate on the treadmill and slowly turn it on. The roller skate moves backwards. Now try and push the roller skate forward. Under normal circumstances, you could push it forward. But if at the very time you attempted to push it forward(much like a propeller screwing into the air), for each tiny, tiny increase in movement you tried to move it forward the conveyor increases it's speed the same amount in the opposite direction, instantly(since the problem said it could match the speed). You would never be able to break the friction of the wheels from the conveyor to get them to start rolling. You would see no visible movement in the conveyor or the wheels. It would be like trying to push two extremely powerful magnets together when the poles were not opposite. You know that feeling? If I can just push them harder, I can overcome the force. But it doesn't work. The harder you push, the harder the conveyor pushes back without moving. The entire will-fly team will drop their jaws, hang their heads in disbelief, tuck their helmets under their arms and walk off the field in disbelief. We are good winners, so we will buy them a root-beer.

 

[Egon]

This aircraft can not fly at the present time. It's been on the darned belt so long that now it's in the shop having the engine rebuilt!

Whatever happens in sealed truck box full of flying sparrows if the truck comes to an full emergency stop??

 

[MossRoad]

This aircraft can not fly at the present time. It's been on the darned belt so long that now it's in the shop having the engine rebuilt!

Whatever happens in sealed truck box full of flying sparrows if the truck comes to an full emergency stop??

I agree. The paperwork on this thing is going to be a nightmare.

 

[patrick_g]

Tom, You assume the elevator was in the Northern hemisphere, a fact not in evidence, but well done!

My FIL was, for a time, a rocket scientist and one of his papers was "Precision Azimuth Allignment of Ballistic Missiles" He worked directly for Werner VB and was liaison to a lot of the "imported German talent." He was a fun guy to banter with about stuff of physics (THE FATHER SCIENCE.) He didn't let you get away with sloppy thinking.

Now MOSS... Making the "will fly" assumtion on the speed of the MCB where the speed of the MCB exactly equals the speed of the plane... Don't get into Zeno's paradox subdividing speeds into infintessimals. The concepts of calculus will handle those. as well as a very simple arrangement described just a tad down the page.

The plane will start moving down the conveyor pretty normally except for the fact that the wheels will always be spinning at twice the airplanes speed, i.e. as the plane accelerates through any given speed however small or large prior to lifting off. The MCB will MATCH the planes speed. The plane will have slightly less takeoff performance since it will have to provide the extra energy to account for the fact that the wheels will accelerate at twice the acceleration of the plane.

If the bearings don't fail, a reasonable assumption, and the tires don't fly apart at twice the normal rotational speed, I'm not sure about the safety margin on aircraft tires, the plane will take off almost as nicely as if the MCB were switched off. For the purposes of this exercise it would be fair to assume the bearings and tires survive as their performance is not the central issue of the mental exercise.

The plane takes off at x MPH while the MCB is producing x MPH and the wheels are rotating at a rate to make 2xMPH.

If you assume low or no friction in the design of the MCB and a low or no weight in its moving parts you could attach a rope with a hook on it to the airplane and through a turning block and drum let the plane's forward motion supply the motion of the MCB. Then there would be no question about the MCB dectecting and responding to infintessimal motions and such distractions.

Every inch forward the plane moves the rope and pulley pulls the MCB an inch in the oppoosite direction at the same speed the plane is moving however small or large (since our rope is a low stretch polyester.)

For safety reasons the rope is attached to the drum like a non-recoil starter rope is attached to the flywheel of an engine so after the end of the rope is reached it disconnects from the drum. after liftoff the pilot releases the mechanical connection of the hook on the plane (like a glider's tow plane.)

Anyway this rope and pulley exactly fits the "WILL FLY" argument while it simplifies the MCB to just CB (not much magic in a rope and pulley.)

The somewhat more fun and considerably more interesting (but a bit less defensible in its enabling assumptions) is the "NO-FLY" contention. If for whatever reason you allow the MCB to run at whatever speed is required to prevent the plane's forward motion and there are no material failures the pllane will run out of gas sitting stationery
on the MCB unless the wind generated by the MCB will lift the plane. If it does lift off it may stall leaving the region where the wind is being generated if the MCB is not sufficiently long.

I assert that even in theory there is no way to attach the rope of the above example to a mechanical mechanism that will cause the MCB to experience an accelerated motion sufficient to cause the plane to remain statioinery since if the plane is stationery there is no motion of the rope.

Pat